To solve $(\sqrt{2} + 1)^4$, we can use the binomial theorem, which states that:

$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$

Here, $a = \sqrt{2}$, $b = 1$, and $n = 4$. Applying the binomial theorem, we have:

$(\sqrt{2} + 1)^4 = \sum_{k=0}^{4} \binom{4}{k} (\sqrt{2})^{4-k} (1)^k$

This simplifies to:

$(\sqrt{2} + 1)^4 = \binom{4}{0} (\sqrt{2})^4 (1)^0 + \binom{4}{1} (\sqrt{2})^3 (1)^1 + \binom{4}{2} (\sqrt{2})^2 (1)^2 + \binom{4}{3} (\sqrt{2})^1 (1)^3 + \binom{4}{4} (\sqrt{2})^0 (1)^4$

Calculating each term separately:

1. $\binom{4}{0} (\sqrt{2})^4 (1)^0 = 1 \cdot 4 \cdot 1 = 4$
2. $\binom{4}{1} (\sqrt{2})^3 (1)^1 = 4 \cdot 2\sqrt{2} \cdot 1 = 8\sqrt{2}$
3. $\binom{4}{2} (\sqrt{2})^2 (1)^2 = 6 \cdot 2 \cdot 1 = 12$
4. $\binom{4}{3} (\sqrt{2})^1 (1)^3 = 4 \cdot \sqrt{2} \cdot 1 = 4\sqrt{2}$
5. $\binom{4}{4} (\sqrt{2})^0 (1)^4 = 1 \cdot 1 \cdot 1 = 1$

Summing these results:

$4 + 8\sqrt{2} + 12 + 4\sqrt{2} + 1 = 17 + 12\sqrt{2}$

So,

$(\sqrt{2} + 1)^4 = 17 + 12\sqrt{2}$

Would you like further details or have any questions?

Here are some related questions for further practice:

1. What is the expansion of $(\sqrt{2} - 1)^4$?
2. How do you simplify $(\sqrt{3} + 1)^3$?
3. What is the value of $(2 + \sqrt{3})^2$?
4. How do you use the binomial theorem to expand $(x + y)^5$?
5. What are the coefficients in the expansion of $(x - 1)^4$?
6. How do you find the middle term in the expansion of $(a + b)^6$?
7. What is the expansion of $(\sqrt{5} + 2)^3$?
8. How do you simplify $(\sqrt{3} - 1)^3$?

Tip: When working with the binomial theorem, it's often helpful to write out a few terms to identify patterns before calculating the entire expansion.